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densk [106]
3 years ago
6

Many freeways have service (or logo) signs that give information on attractions, camping, lodging, food, and gas services prior

to off-ramps. These signs typically do not provide information on distances. The article Evaluation of Adding Distance Information to Freeway-Specific Service (Logo) Signs (J. of Transp. Engr., 2011; 78-888) reported that in one investigation, six sites along Virginia interstate high ways where service signs are posted were selected. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of crashes per year before and after the sign changes were as follows:
Before: 15 26 66 115 62 64
After: 16 24 42 80 78 73
The cited article included the statement A paired t test was performed to determine whether there was any change in the mean number of crashes before and after the addition of distance information on the signs. Carry out such a test. [Note: The relevant normal probability plot shows a substantial linear pattern.] a seventh site were to be randomly selected among bearing service signs, between what values predict the difference in number of crashes to lie?
Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

H0 : μd = 0

H1 : μd ≠ 0

Test statistic = 0.6687 ;

Pvalue = 0.7482 ;

Fail to reject H0.

Step-by-step explanation:

H0 : μd = 0

H1 : μd ≠ 0

Given the data:

Before: 15 26 66 115 62 64

After: 16 24 42 80 78 73

Difference = -1 2 24 35 -18 -9

Mean difference, d ; Σd / n

d = Σx / n = ((-1) + 2 + 24 + 35 + (-18) + (-9))

d = Σx / n = 33 / 6 = 5.5

Test statistic = (d / std / sqrt(n))

std = sample standard deviation = 20.146

Test statistic = 5.5 ÷ (20.146/sqrt(6))

Test statistic = 0.6687

The Pvalue :

P(Z < 0.6687) = 0.7482

At α = 0.05

Pvalue > α ; Hence we fail to reject H0

The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.

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3 years ago
NO LINKS!!! What is the equation of these two graphs?​
S_A_V [24]

Answer:

\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}

\textsf{2.} \quad y=-|x+1|+5

Step-by-step explanation:

<h3><u>Question 1</u></h3>


<u>Method 1 - modelling as a quadratic with restricted domain</u>


Assuming that the points given on the graph are points that the <u>curve passes through</u>, the curve can be modeled as a quadratic with a limited domain.  Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

y=ax^2+bx+c

Given points:

  • (-4, 2)
  • (-1, 4)
  • (4, 6)

Substitute the given points into the equation to create 3 equations:

<u>Equation 1  (-4, 2)</u>

\implies a(-4)^2+b(-4)+c=2

\implies 16a-4b+c=2

<u>Equation 2  (-1, 4)</u>

\implies a(-1)^2+b(-1)+c=4

\implies a-b+c=4


<u>Equation 3  (4, 6)</u>

\implies a(4)^2+b(4)+c=6

\implies 16a+4b+c=6

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

\implies (16a+4b+c)-(16a-4b+c)=6-2

\implies 8b=4

\implies b=\dfrac{4}{8}

\implies b=\dfrac{1}{2}

Subtract Equation 2 from Equation 3 to eliminate variable c:

\implies (16a+4b+c)-(a-b+c)=6-4

\implies 15a+5b=2

\implies 15a=2-5b

\implies a=\dfrac{2-5b}{15}

Substitute found value of b into the expression for a and solve for a:

\implies a=\dfrac{2-5(\frac{1}{2})}{15}

\implies a=-\dfrac{1}{30}

Substitute found values of a and b into Equation 2 and solve for c:

\implies a-b+c=4

\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4

\implies c=\dfrac{68}{15}

Therefore, the equation of the graph is:

y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}

\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}

<u>Method 2 - modelling as a square root function</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, and the x-intercept should be included, we can model this curve as a <u>square root function</u>.

Given points:

  • (-4, 2)
  • (-1, 4)
  • (4, 6)
  • (0, -5)

The parent function is:

y=\sqrt{x}

Translated 5 units left so that the x-intercept is (0, -5):

\implies y=\sqrt{x+5}

The curve is stretched vertically, so:

\implies y=a\sqrt{x+5} \quad \textsf{(where a is some constant)}

To find a, substitute the coordinates of the given points:

\implies a\sqrt{-4+5}=2

\implies a=2

\implies a\sqrt{-1+5}=4

\implies 2a=4

\implies a=2

\implies a\sqrt{4+5}=6

\implies 3a=6

\implies a=2

As the value of a is the same for all points, the equation of the line is:

y=2\sqrt{x+5}

<h3><u>Question 2</u></h3>

<u>Vertex form of an absolute value function</u>

f(x)=a|x-h|+k

where:

  • (h, k) is the vertex
  • a is some constant

From inspection of the given graph:

  • vertex = (-1, 5)
  • point on graph = (0, 4)

Substitute the given values into the function and solve for a:

\implies a|0-(-1)|+5=4

\implies a+5=4

\implies a=-1

Substituting the given vertex and the found value of a into the function, the equation of the graph is:

y=-|x+1|+5

7 0
2 years ago
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